7.1 INTRODUCTION TO PERIODIC FUNCTIONS

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1 7.1 INTRODUCTION TO PERIODIC FUNCTIONS

2 *SECTION: 6.1 DCP List: periodic functions period midline amplitude Pg 247- LECTURE EXAMPLES: Ferris wheel, 14,16,20, eplain 23, 28, 32 *SECTION: 6.2 DCP List: unit circle sine cosine coordinates negative rotation Pg 255- LECTURE EXAMPLES: 6, 22, 26,28 *SECTION: 6.3 DCP List: radian relationship to degrees arc length Pg 261- LECTURE EXAMPLES: 6,12,15h, 24,36,40 *SECTION: 6.4 DCP List: sine and cosine graphs amplitude Pg 267- LECTURE EXAMPLES: 2,14,20,23d,30 *SECTION: 6.5 DCP List: sinusoidal functions transformations phase shift Pg 275-LECTURE EXAMPLES: 2,8,16,20,34, 41,42

3 *SECTION: 6.6 (small review of trigonometry) minimal coverage. must know. DCP List tangent secant cosecant cotangent Pythagorean identities check your knowledge of reciprocals, si trig functions and evaluation Pg 283-LECTURE EXAMPLES: 14,18, week 22,32 *SECTION: 6.7 DCP List: inverse trigonometric functions solving trigonometric equations reference angle Pg 292-LECTURE EXAMPLES: 6,16,32,42,52,58, (54)notation with eponent of -1 *Chapter Si Review eercises and problems pg : 1,3,18,19,21,27,29,31,35,41,43,49,51,55,59 weekend 57,62,64 LECTURE EXAMPLES week 44,48,50,56 CHECK YOUR UNDERSTANDING page : 1-11odd,14,19,22,24,27,29,33,38, 41,42,43,45,47,48,51,55,59, 63,67,71,73,76,77,80,85,87

4 London Eye 135-metre (443 ft) tall. It rotates at 26 cm (10 in) per second (about 0.9 km/h or 0.6 mph) so that one revolution takes about 45 minutes.

6 Singapore Flyer Each of the 28 air-conditioned Capsules is capable of holding 28 passengers, and a complete rotation of the wheel takes Approimately 37 minutes. Constructed in Described as an observation wheel, that reaches 42 stories high, with a total height of 165 m (541 ft)

7 Ferris Wheel Height As a Function of Time The London Eye Ferris Wheel measures 450 feet in diameter and turns continuously, completing a single rotation once every 30 minutes. Suppose you hop on the London Eye Ferris wheel at time t = 0 and ride it for two full turns. Let f(t) be your height above the ground, measured in feet as a function of t, the number of minutes you have been riding. We can figure out some values of f(t). Since the speed of the rotation is constant, you are at the top 15 minutes after boarding and one-quarter of the way up at 7.5 minutes and 22.5 minutes after boarding. Then you are back at the bottom after 30 minutes and the process continues. Values of f(t), your height above the ground t minutes after boarding the wheel t (minutes) f(t) (feet) t (minutes) f(t) (feet)

8 Graphing the Ferris Wheel Function Values of f(t), your height above the ground t minutes after boarding the wheel t (minutes) f(t) (feet) t (minutes) f(t) (feet) Notice that the values of f(t) in the table begin repeating after 30 minutes since the second turn is just like the first turn, ecept that it happens 30 minutes later. If you ride the wheel for more full turns, the values of f(t) continue to repeat at 30-minute intervals. We plot this data and fill in the blank spaces. 450 y (feet) t (minutes) Functions Modeling Change:

10 Periodic Functions: Period, Midline, and Amplitude The Ferris Wheel function, f, is said to be periodic, because its values repeat on a regular interval or period. In the figure, the period is indicated by the horizontal gap between the first two peaks. The dashed horizontal line is the midline of the graph of f. The vertical distance shown between the first peak and the midline is called the amplitude. y (feet) 450 Amplitude: Radius of wheel is 225 ft 225 Period: One rotation takes 30 minutes Midline: Wheel s hub is 225 ft above ground y = The graph of y = f(t) showing the amplitude, period, and midline t (minutes) Functions Modeling Change:

11 The Singapore Flyer in, 37 minutes, completes a rotation that reaches a total height of 165 m. Let f(t) be your height above the ground, measured in feet as a function of t, the number of minutes you have been riding. Complete the table and sketch f(t). Label the period, amplitude and midline. t (minutes) 0 f(t) (feet)

12 7.2 THE SINE AND COSINE FUNCTIONS

13 Using Angles to Measure Position On a Circle Conventions For Working With Angles We measure angles with respect to the horizontal, not the vertical, so that 0 describes the 3 o clock position. Positive angles are measured in the counterclockwise direction, negative angles in the clockwise direction. Large angles (greater than 360 or less than 360 ) rotate around a circle more than once.

14 Height on the Ferris Wheel as a Function of Angle Since we can measure position on the Ferris wheel using angles, we see that the height above the ground is a function of the angle position on the wheel. We can rewrite our table giving heights as a function of angle, instead of time. Your height above ground, y, as a function of the angle turned through by the wheel θ (degrees) f(t) (feet) θ (degrees) f(t) (feet) Recall how the y-values repeat every 30 minutes. Similarly, in the Table, the values of y repeat every 360. In both cases, the y-values repeat every time the wheel completes one full revolution.

15 The Unit Circle When we studied quadratic functions, we transformed a special starting function y = 2 to get other quadratic functions. Similarly, we begin here with a special starting circle. This is the unit circle, the circle of radius one centered at the origin. The unit circle gets its name from the fact that its radius measures eactly one unit. (-1,0) (0,1) (0,0) (0,-1) y Radius is 1 θ Origin (1,0) P = (, y) Functions Modeling Change:

16 CONVERT TO RADIAN θ (degrees) RADIAN f(t) (feet) θ (degrees) RADIAN f(t) (feet)

17 The Sine and Cosine Functions If θ is an angle in a right triangle (other than the right angle), sin θ = Opposite/Hypotenuse cos θ = Adjacent/Hypotenuse Consider the right triangle formed in the circle of radius r in the Figure: sin θ = y/r cos θ = /r In Right Triangles y Hypotenuse: r = Radius θ Adjacent: P = (, y) Opposite: y

18 The Sine and Cosine Functions Suppose P = (, y) in the figure is the point on the unit circle specified by the angle θ. We define the functions, cosine of θ, or cos θ, and sine of θ, or sin θ, by cos θ = and sin θ = y. In other words, cos θ is the -coordinate of the point P; and sin θ is the y-coordinate. y y = sin(θ) θ = cos(θ) Unit Circle P = (, y)

19 sin θ = y/r cos θ = /r tan θ = / csc θ = / sec θ = / cot θ = / The Functions In Right Triangles y Hypotenuse: r = Radius θ Adjacent: P = (, y) Opposite: y

20 The Unit Circle Find the values of sin θ and cos θ (0,1) y θ sin θ cos θ 0 π/2 (-1,0) (0,0) Radius is 1 θ (1,0) P = (, y) π Origin 3π/2 2π (0,-1)

21 7.3 GRAPHS OF SINE AND COSINE

22 Tabulating and Graphing Values of Sine and Cosine (convert to radian) Values of sin θ and cos θ for 0 θ < 2π θ sin θ cos θ θ sin θ cos θ θ sin θ cos θ θ sin θ cos θ 0 π/6 π/4 π/3 1 y y = cos θ θ in degrees y = sin θ -1 Functions Modeling Change:

23 Tabulating and Graphing Values of Sine and Cosine Values of sin θ and cos θ for for 0 θ < 2π θ sin θ cos θ θ sin θ cos θ θ sin θ cos θ θ sin θ cos θ π/ π/ π/ y y = cos θ θ in degrees y = sin θ -1 Functions Modeling Change:

24 Properties of Sine and Cosine Properties of the sine and cosine functions that are apparent from the graph include: Domain: All values of θ, since any angle, positive or negative, specifies a point on the unit circle. Range: Since values of the sine and cosine are coordinates of points on the unit circle, they lie between 1 and 1. So the range of the sine and cosine are 1 sin θ 1 and 1 cos θ 1. Odd/Even Symmetry: The sine function is odd and the cosine function is even: sin( θ) = sin θ and cos( θ) = cos θ. Period: Both sine and cosine are periodic functions, because the values repeat regularly. The smallest interval over which the function values repeat here 360 is called the period. We have sin(θ ) = sin θ and cos(θ ) = cos θ. Functions Modeling Change:

25 Periodic Functions A function f is periodic if its values repeat at regular intervals. Then if the graph of f is shifted horizontally by c units, for some constant c, the new graph is identical to the original graph. In function notation, periodic means that, for all t in the domain of f, f(t + c) = f(t). The smallest positive constant c for which this relationship holds for all values of t is called the period of f.

26 Amplitude and Midline (convert to radian) Eample 1 Compare the graph of y = sin t to the graphs of y = 2 sin t and y = 0.5 sin t, for 0 t 2π. How are these graphs similar? How are they different? What are their amplitudes? Solution: The graphs are shown. The amplitude of y = sin t is 1, the amplitude of y = 2 sin t is 2 and the amplitude of y = 0.5 sin t is 0.5. The graph of y = 0.5 sin t is upside-down relative to y = sin t. These observations are consistent with the fact that the constant A in the equation y = A sin t stretches or shrinks the graph vertically, and reflects it about the t-ais if A is negative. The amplitude of the function is A. y y = 2sin t, amplitude= t y = -0.5 sin t, amplitude=0.5 y = sin t, amplitude=1 The t ais is the midline for all three functions. Functions Modeling Change:

27 Eample Midlines and Vertical Shifts Compare the graph of y = cos t to the graphs of y = cos t + 3 and y = cos t - 2, for 0 t 2π. How are these graphs similar? How are they different? What are their midlines? Solution: The graphs are shown. The midline of y = cos t is the t ais (y = 0), the midline of y = cos t + 3 is y = 3 and the midline of y = cos t - 2 is y = -2. Recall that when the function f(t) is shifted vertically by a distance k, the new function is f(t) + k. Similarly, the midline is shifted vertically by that same distance k. Generalizing, we conclude that the graphs of y = sin t + k and y = cos t + k have midlines y = k. y y = cos t + 3, midline: y=3 y = cos t, midline: y= t y = cos t 2, midline: y=-2 The t ais is the midline for all three functions. Functions Modeling Change:

28 Coordinates of a Point on a Circle of Radius r The coordinates of the point P on the unit circle y in the figure are given by = cos θ and Q =(, y) y = sin θ. The coordinates (, y) of the point Q in the r 1 θ P = r cosθ y = r sinθ figure are given by = r cos θ and y = r sin θ.

29 Coordinates of Points on a Circle of Radius r = 5 (convert to radian) Eample 2 Find the coordinates of the points A, B, and C in the Figure to three decimal places. Solution With r = 5, the coordinates of point A are given by = 5 cos 130 = 5( ) = 3.214, y = 5 sin 130 = 5(0.766) = Point B corresponds to an angle of 70, (angle is measured clockwise), so B has coordinates = 5 cos( 70 ) = 5(0.342) = 1.710, y = 5 sin( 70 ) = 5( ) = For point C, we must first calculate the corresponding angle, since the 10 is not measured from the positive -ais. The angle we want is = 190, so = 5 cos(190 ) = 5( ) = 4.924, y = 5 sin(190 ) = 5( ) = C A 10 r = 5 y 130 B 70 Functions Modeling Change:

30 Height on the Ferris Wheel Eample 4 The Ferris wheel has a radius of 225 feet. Find your height above the ground as a function of the angle θ measured from the 3 o clock position. What is your height when θ = 60? when θ = 150? Solution as a Function of Angle (0,0) y θ P = (, y) At θ = 60, h(t) = sin 60 = ft. At θ = 150, h(t) = sin 150 = ft. Height = y = sin θ 225 y = 225 sinθ 225 Functions Modeling Change:

31 Height on the Ferris Wheel as a Function of Angle Eample 5 Graph the Ferris wheel function giving your height, h = f(θ), in feet, above ground as a function of the angle θ : f(θ) = sin θ. What are the period, midline, and amplitude? Solution: A calculator gives the values to graph f(θ). The period of this function is 2π, because 2π is one full rotation, so the function repeats every 360. The midline is h = 225 feet, since the values of h oscillate about this value. The amplitude is also 225 feet, since the maimum value of h is 450 feet. Amplitude = 225 ft h, height (feet) 450 Period: 2π f (θ) = sin θ 225 Midline: y = 225 Θ (degrees) Functions Modeling Change:

32 7.4 THE TANGENT FUNCTION

33 The Tangent Function Suppose P = (, y) in the figure is the point on the unit circle specified by the angle θ. We define the function, tangent of θ, or tan θ by tan θ = y / for 0. y θ 1 y P = (, y) Since = cos θ and y = sin θ, we see that tan θ = sin θ/cos θ for cos θ 0.

34 The Tangent Function in Right Triangles If θ is an angle in a right triangle (other than the right angle), a Opposite tan b Adjacent c a θ b

35 The Tangent Function in Right Triangles Eample 3 The grade of a road is calculated from its vertical rise per 100 feet. For instance, a road that rises 8 ft in every one hundred feet has a grade of Grade = 8 ft/100 ft = 8%. Suppose a road climbs at an angle of 6 to the horizontal. What is its grade? Solution: From the figure, we see that tan 6 = /100, so, using a calculator, = 100 tan 6 = Thus, the road rises ft every 100 feet, so its grade is 10.51/100 = 10.51% ft A road rising at an angle of 6 (not to scale)

36 Interpreting the Tangent Function as Slope We can think about the tangent function in terms of slope. In the Figure, the line passing from the origin through P has Slope y y 0 0 Slope In words, tan θ is the slope of the line passing through the origin and point P. y y so tan θ. (0, 0) P = (, y) y θ Line has slope y/ = tan θ

37 Graphing the Tangent Function (convert to radian) For values of θ between 180 and 360, observe that tan(θ ) = tan θ, because the angles θ and θ determine the same line through the origin, and hence the same slope. Thus, y = tan θ has period 180. Since the tangent is not defined when the -coordinate of P is zero, the graph of the tangent function has a vertical asymptote at θ = 270, 90, 90, 270, etc. Graph of the tangent function Θ (degrees)

38 7.5 RIGHT TRIANGLES: INVERSE TRIGONOMETRIC FUNCTIONS

39 The Inverse Sine Function For 0 1: arcsin = sin 1 which means The angle in a right triangle whose sine is. Eample 2 Use the inverse sine function to find the angles in the figure. θ 5 φ Solution: (change the calculator mode) Using our calculator s inverse sine function: sin θ = 3/5 = 0.6 so θ = sin 1 (0.6) = sin φ = 4/5 = 0.8 so φ = sin 1 (0.8) = These values agree with the ones found in Eample

40 The Inverse Tangent Function arctan = tan 1 = The angle in a right triangle whose sine is. Eample 3 The grade of a road is 5.8%. What angle does the road make with the horizontal? Solution: Since the grade is 5.8%, the road climbs 5.8 feet for 100 feet; see the figure. We see that tan θ = 5.8/100 = So θ = tan 1 (0.058) = ft θ 100 ft A road rising at a grade of 5.8% (not to scale)

Summary of Inverse Trigonometric Functions We define: the arc sine or inverse sine function as arcsin = sin 1 = The angle in a right triangle whose sine is the arc cosine or inverse cosine function

41 Summary of Inverse Trigonometric Functions We define: the arc sine or inverse sine function as arcsin = sin 1 = The angle in a right triangle whose sine is the arc cosine or inverse cosine function as arccos = cos 1 = The angle in a right triangle whose cosine is the arc tangent or inverse tangent function as arctan = tan 1 = The angle in a right triangle whose tangent is. This means that for an angle θ in a right triangle (other than the right angle), sin θ = means θ = sin 1 cos θ = means θ = cos 1 tan θ = means θ = tan 1.

42 Eamples

43 A company s sales are seasonal with the peak in mid-december and the lowest point in mid-june. The company makes $100,000 in sales in December, and only $20,000 in June. (a) Find a trigonometric function, s = f(t), representing sales at time t months after mid-january. (b) What would you epect the sales to be for mid-april? (c) Find the t-values for which s = 60,000. Interpret your answer

44 1) Sketch the terminal ray, label the reference value and find the eact value without a calculator csc ) Evaluate: csc 1 csc 35 3

45 KNOW THESE

46 3) Solve : 3sin 2 3sin 4 3 2sin 4) prove identity : 1 cos 2 tan 2cos sin

47 Simplify ) tan )(1 tan (1 1)cos 2 y y y y tan tan 1 tan tan sin sin cos cos sin cos cos sin 3) cos 1 tan sin 1 cos 2)

48 Graph each to see if any two epressions appear to be identical. Prove any identities you find. tan 2 sin 2 (tan 2 )(sin 2 ) tan 2 sin 2

49 A utility company serves two different cities. Let P1 be the power requirement in megawatts (mw) for City 1 and P2 be the requirement for City 2. Both P1 and P2 are functions of t, the number of hours elapsed since midnight. Suppose P1 and P2 are given by the following formulas:

50 (a) Describe the power requirements of each city in words. (b) What is the maimum total power the utility company must be prepared to provide?

51 7.6 NON-RIGHT TRIANGLES Done in previous course or MAT 109

52 The Law of Cosines Law of Cosines: For a triangle with sides a, b, c, and angle C opposite side c, we have c 2 = a 2 + b 2 2ab cos C A b c C a B

53 Proof of the Law of Cosines Applying the Pythagorean theorem to the right-hand right triangle: (a ) 2 + h 2 = c 2 or a 2 2a h 2 = c 2. Applying the Pythagorean theorem to the left-hand triangle, we get 2 + h 2 = b 2. Substituting this result into the previous equation gives a 2 2a + ( 2 + h 2 ) = a 2 2a + b 2 = c 2. But cos C = /b, so = b cos C. This gives the Law of Cosines: a 2 + b 2 2ab cos C = c 2. A b h c C a a Triangle used to derive the Law of Cosines B

54 Application of the Law of Cosines Eample 1 A person leaves her home and walks 5 miles due east and then 3 miles northeast. How far has she walked? How far away from home is she? Solution She has walked = 8 miles in total. One side of the triangle is 5 miles long, while the second side is 3 miles long 3 and forms an angle of 135 Home with the first. This is because when the person turns northeast, she turns through an angle of 45. Thus, we know two sides of this triangle, 5 and 3, and the angle between them, which is 135. To find her distance from home, we find the third side, using the Law of Cosines: cos Notice that this is less than 8 miles, the total distance she walked., Destination miles

55 The Law of Sines Law of Sines: For a triangle with sides a, b, c opposite angles A, B, C respectively: sin A a sin B b A sin C c b c C a B

56 Proof of the Law of Sines We derive the Law of Sines using the same triangle as in the proof of the Law of Cosines. Since sin C = h/b and sin B = h/c, we have h = b sin C and h = c sin B. This means that b sin C = c sin B and sin B sin C A similar type of argument (Problem 42) shows that. b c A sin A sin B a b b h c C Triangle used to derive the Law of Sines a B

57 Application of the Law of Sines Eample 3 An aerial tram starts at a point one half mile from the base of a mountain whose face has a 60 angle of elevation. (See figure.) The tram ascends at an angle of 20.What is the length of the cable from T to A? A Solution: T a = 0.5 mile C The Law of Cosines does not help us here because we only know the length of one side of the triangle. We do however know two angles in this diagram and can determine the third. Thus, we can use the Law of Sines: sina/a = sinc/c or sin 40 /0.5 = sin 120 /c So c = 0.5 (sin 120 /sin 40 ) = Therefore, the cable from T to A is miles. c 40 Functions Modeling Change:

58 When to Use the Laws of Cosines and Sines When two sides of a triangle and the angle between them are known the Law of Cosines is useful. It is also useful if all three sides of a triangle are known. The Law of Sines is useful when we know a side and the angle opposite it and one other angle or one other side. The Ambiguous Case: There is a drawback to using the Law of Sines for finding angles. The problem is that the Law of Sines does not tell us the angle, but only its sine, and there are two angles between 0 and 180 with a given sine. For eample, if the sine of an angle is 1/2, the angle may be either 30 or 150.

7.1 INTRODUCTION TO PERIODIC FUNCTIONS

7.1 INTRODUCTION TO PERIODIC FUNCTIONS Ferris Wheel Height As a Function of Time The London Eye Ferris Wheel measures 450 feet in diameter and turns continuously, completing a single rotation once every